RBSE Solutions for Class 7 Maths Chapter 10 Construction of Triangles In Text Exercise is part of RBSE Solutions for Class 7 Maths. Here we have given Rajasthan Board RBSE Class 7 Maths Chapter 10 Construction of Triangles In Text Exercise.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 7 |

Subject |
Maths |

Chapter |
Chapter 10 |

Chapter Name |
Construction of Triangles |

Exercise |
In Text Exercise |

Number of Questions |
11 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 7 Maths Chapter 10 Construction of Triangles In Text Exercise

**Do and Learn**

**(Page 123)**

Question 1

Construct a ∆XYZ in which XY = 4.5 cm, YZ = 5 cm and ZX = 6 cm.

Solution:

Steps of construction:

1. Draw a line (RBSESolutions.com) segment of length XY = 4.5 cm.

2. Draw an arc of radius 6 cm from point X

3. Draw an arc of radius 5 cm from point Y, which cuts the prior arc at point Z.

4. Join XZ and YZ. Thus AXYZ is required triangle.

Question 2

Construct an equilateral (RBSESolutions.com) triangle of side 5.5 cm.

Solution:

Steps of construction:

1. Draw a line segment of length AB = 5.5 cm.

2. Draw an arc of radius 5.5 cm from A and B each which intersect each other at point C.

3. Join AC and BC.

Thus ∆ABC is required equilateral triangle.

Question 3

Draw a ∆PQR with PQ = 4 cm, QR = 3. 5 cm and PR = 4. What type (RBSESolutions.com) of triangle is this?

Solution:

Steps of construction:

1. Draw a line segment PQ = 4.

2. Draw (RBSESolutions.com) an arc of radius 4 cm from point P.

3. Draw another arc of radius 3.5 cm which intersects prior arc at point R.

4. Join PR and QR.

Thus ∆PQR is required tirangle.

also ∵ PQ = QR = 4 cm.

∴ Thus ∆PQR is an isosceles triangle.

**(Page 125)**

Question 1

(i) Construct a ∆DEF such (RBSESolutions.com) that DE = 5 cm, DF = 3 cm and ∠EDF = 90°.

Solution:

Steps of construction:

1. Draw a line segment DE = 5 cm.

2. Make an angle of 90° at point D.

3. Make an arc of 3 cm from point D which cuts the line of angle at point F.

4. Join EF.

Thus ∆DEF is required triangle.

Q. ii. Construct an isosceles triangle is (RBSESolutions.com) which the lengths of each of its equal side is 6.5 cm and angle between them is 110°.

Solution:

Steps of construction:

1. Draw a line segment BC = 6.5 cm

2. Draw a ray A from point C, making an angle of 110°.

3. Make an arc of 6.5 cm from point B so that it (RBSESolutions.com) intersects the ray BX at point A.

4. Join AC

Thus ΔABC is required triangle.

Q. iii. Construct ∆ABC with BC = 7.5 cm, AC = 5 cm and ∠C = 60°.

Solution:

Steps of construction:

1. Draw a line segment BC= 7.5 cm

2. Draw a ray from point C, making an angle of 60°.

3. Make an (RBSESolutions.com) arc of 5 cm from point C which inctrsects the ray CX at point A.

4. Join A to B.

Thus ∆ABC is required triangle.

**(Page 126)**

Question 1

Construct ∆ABC, given ∠A = 60°, ∠B = 30° and AB = 5.8.

Solution:

Steps of construction:

1. Draw a line (RBSESolutions.com) segment AB = 5.8 cm

2. Make an angle of ∠BAX = 60° at point A.

3. Make an angle of ∠ABY = 30° at point B.

4. AX and BY intersect each other at point C.

Thus ∆ABC is required triangle.

Question 2

Construct ∆PQR if PQ = 5 cm, ∠PQR = 105° and ∠QRP = 40°.

(Hint : Recall angle sum property of a triangle)

Solution:

Here PQ = 5 cm ∠Q = 105°, ∠R = 40°

∴ ∠P = ?

we know that

∠P + ∠Q + ∠R = 180°

⇒ ∠P + 105° + 40° = 180°

⇒ ∠P + 145° = 180°

⇒ ∠P = 180° – 145° = 35°

PQ = 5 cm ∠P = 35° and ∠Q= 105°

Now ∆PQR can (RBSESolutions.com) be constructed.

Steps of construction :

1. Draw a line segment PQ = 5 cm

2. Draw a ray PX making an angle of 35° at P.

3. Draw a ray QY making an angle of 105° at Q.

4. PX and QY intersect at R.

Thus ∆PQR is a required triangle.

Question 3

Examine whether you can (RBSESolutions.com) construct ∆DEF such that EF = 7.2 cm, ∠E = 110° and ∠F = 80°, justify your answer.

Solution:

Given that

∠E = 110° and ∠F=80°

∵ ∠E + ∠F = 110°+ 80°= 190°

∵ Sum of only two angles is 190° which is greater than 180° (sum of all three angles)

∴ Construction of ∆DEF is not possible.

**(Page 128)**

Question 1

Construct the right angled (RBSESolutions.com) triangle ∆PQR, where ∠Q = 90°, QR = 8 cm and PR = 10 cm.

Solution:

Steps of construction :

1. Draw QR = 8 cm.

2. Make ∠XQR = 90° at point Q.

3. Make an arc of 10 cm from point R which cuts QX at point P.

4. Join RP.

Thus ∆PQR is required triangle

Question 2

Construct a right-angled triangle whose (RBSESolutions.com) hypotenuse is 6 cm long and one of the legs is 4 cm long.

Solution:

Steps of construction :

1. Draw QR = 4 cm

2. Make ∠XQR = 90° at point Q.

3. Make an arc of 6 cm (RBSESolutions.com) from point Q which cuts QX at point P.

4. Join RP

Thus ∆PQR is required triangle.

We hope the RBSE Solutions for Class 7 Maths Chapter 10 Construction of Triangles In Text Exercise will help you. If you have any query regarding Rajasthan Board RBSE Class 7 Maths Chapter 10 Construction of Triangles In Text Exercise, drop a comment below and we will get back to you at the earliest.